Self-induction of Wires. 177 



stants concerned, then, to ensure conjugacy, we require 



a Q = 0, «! = 0, a 2 = 0, &c. . . . (5c) 



separately ; and if these a's cannot all vanish together we 

 cannot have conjugacy. 



What C may be then depends only upon the initial state of 

 the system in subsiding, or upon other impressed forces that 

 we have nothing to do with. As depending upon the initial 

 state, the solution is 



= 2AeJ"i (6c) 



the summation being with respect to the p's which are the 

 roots of F(p)=0,p being put for d/dt in F; and the A 

 belonging to a certain p is to be obtained by the conjugate 

 property of the equality of the mutual electric to the mutual 

 magnetic energy of the normal systems of any pair of p's. 



As depending upon e, the impressed force in the conductor 

 which is to be conjugate to the one in which the current is C, 

 let e be zero before time t = 0, and constant after. Then, 

 by (3c), 



_ /{d/dt)e _ %f f(p)e 



^~F{dfdT)- Z ^p¥' [l£ ' 



= °o- 2 ^^ ( 7 *) 



if C is the final steady current, and F' = dF/dj?, the summa- 

 tion being with respect to the p's. 



If there is a resistance-balance, a =0, C = 0, and 



C = 2^V (8c) 



Now, subject to (4c), calculate the integral transient cur- 

 rent : — 



J, 



f(p)e 



-v 2 F n 



Cdt = % 

 h -F 



= value of f{p)e/p¥(p) whenp = 0, 



= <h/?o, (9c) 



if F is thep = value of F. If then a Y = also, we prove 

 that the integral transient current is zero. 

 Supposing both a =0, ^=0, then 



(J _ g Cl 2P 2 + - . » € pt. 



therefore p$' ' ' 



